The total mechanical energy of an orbiting body is the sum of kinetic energy and gravitational potential energy:
\[
\boxed{E=K+U}.
\]
In Newtonian gravity, with the zero of gravitational potential energy chosen at infinity,
\[
\boxed{U=-\frac{GMm}{r}}.
\]
The negative sign is important. It shows that an orbiting object at finite distance is gravitationally bound unless it has enough kinetic energy to reach \(E\ge0\).
1. Kinetic energy
The kinetic energy of an orbiting object of mass \(m\) and speed \(v\) is
\[
\boxed{K=\frac12mv^2}.
\]
Therefore, at an instant when the object is at radius \(r\), the total mechanical energy is
\[
\boxed{
E=\frac12mv^2-\frac{GMm}{r}
}.
\]
2. Specific mechanical energy
It is often useful to divide total energy by the orbiting mass \(m\). This gives the specific mechanical energy:
\[
\boxed{
\varepsilon=\frac{E}{m}
}.
\]
Substituting \(E=K+U\),
\[
\boxed{
\varepsilon=\frac{v^2}{2}-\frac{GM}{r}
}.
\]
Specific energy is measured in joules per kilogram, which is equivalent to \(\mathrm{m^2/s^2}\).
3. Circular orbit energy
For a circular orbit, gravity provides the centripetal force:
\[
\frac{GMm}{r^2}=\frac{mv^2}{r}.
\]
Cancel \(m\) and one factor of \(r\):
\[
v^2=\frac{GM}{r}.
\]
The circular-orbit kinetic energy becomes
\[
K=\frac12mv^2
=
\frac12m\left(\frac{GM}{r}\right)
=
\boxed{\frac{GMm}{2r}}.
\]
The gravitational potential energy is
\[
U=-\frac{GMm}{r}.
\]
Therefore,
\[
E=K+U
=
\frac{GMm}{2r}
-
\frac{GMm}{r}.
\]
\[
\boxed{
E=-\frac{GMm}{2r}
}.
\]
For a circular orbit,
\[
K=-\frac12U,
\qquad
E=\frac12U=-K.
\]
4. Bound elliptical orbit energy
For any bound Keplerian orbit, the total mechanical energy depends only on the semi-major axis \(a\), not on eccentricity:
\[
\boxed{
E=-\frac{GMm}{2a}
}.
\]
The circular orbit is the special case where \(a=r\).
This means two elliptical orbits with the same semi-major axis have the same total energy even if one is more stretched than the other.
5. Why negative energy means a closed orbit
The escape point is defined by
\[
E=0.
\]
If
\[
E<0,
\]
the orbiting object does not have enough energy to reach infinity with nonzero separation speed. It is bound.
If
\[
E=0,
\]
the object is at the parabolic escape threshold.
If
\[
E>0,
\]
the object is unbound and can escape with remaining speed at infinity.
6. Escape speed from energy
Starting from the instantaneous energy equation,
\[
E=\frac12mv^2-\frac{GMm}{r}.
\]
Set \(E=0\) for just escape:
\[
0=\frac12mv_{\mathrm{esc}}^2-\frac{GMm}{r}.
\]
Rearranging gives
\[
\frac12mv_{\mathrm{esc}}^2=\frac{GMm}{r}.
\]
Cancel \(m\):
\[
\frac12v_{\mathrm{esc}}^2=\frac{GM}{r}.
\]
Therefore,
\[
\boxed{
v_{\mathrm{esc}}=\sqrt{\frac{2GM}{r}}
}.
\]
For a circular orbit at the same radius,
\[
v_{\mathrm{circ}}=\sqrt{\frac{GM}{r}}.
\]
So
\[
\boxed{
v_{\mathrm{esc}}=\sqrt{2}\,v_{\mathrm{circ}}
}.
\]
7. Energy needed to escape from a bound orbit
A bound orbit has \(E<0\). To reach the escape threshold,
\[
E_{\mathrm{final}}=0.
\]
The energy that must be added is
\[
\Delta E_{\mathrm{escape}}
=
0-E
=
\boxed{-E}.
\]
For a circular orbit,
\[
\Delta E_{\mathrm{escape}}
=
\frac{GMm}{2r}.
\]
This equals the circular-orbit kinetic energy at that radius.
8. Worked example: satellite in circular Earth orbit
Suppose a satellite of mass \(m=1000\ \mathrm{kg}\) is in a circular orbit at radius
\[
r=6.771\times10^6\ \mathrm{m}.
\]
Use
\[
G=6.67430\times10^{-11}\ \mathrm{N\,m^2/kg^2},
\qquad
M_{\oplus}=5.9722\times10^{24}\ \mathrm{kg}.
\]
The total mechanical energy is
\[
E=-\frac{GMm}{2r}.
\]
Substitute:
\[
E=
-\frac{
(6.67430\times10^{-11})
(5.9722\times10^{24})
(1000)
}
{2(6.771\times10^6)}.
\]
\[
E\approx-2.94\times10^{10}\ \mathrm{J}.
\]
Therefore,
\[
\boxed{
E\approx-29.4\ \mathrm{GJ}
}.
\]
The negative sign means the satellite is gravitationally bound.
9. Orbit type by energy sign
| Energy sign |
Orbit type |
Meaning |
| \(E<0\) |
Bound ellipse or circle |
The object stays gravitationally bound unless energy is added |
| \(E=0\) |
Parabolic escape |
The object just escapes with zero speed at infinity |
| \(E>0\) |
Hyperbolic/unbound trajectory |
The object escapes with leftover speed at infinity |
10. Summary formulas
| Goal |
Formula |
Use |
| Instantaneous kinetic energy |
\(K=\frac12mv^2\) |
Energy due to speed |
| Gravitational potential energy |
\(U=-GMm/r\) |
Energy relative to zero at infinity |
| Total mechanical energy |
\(E=K+U\) |
Energy used to classify bound or unbound motion |
| Circular orbit energy |
\(E=-GMm/(2r)\) |
Total energy for a circular orbit |
| Bound orbit energy |
\(E=-GMm/(2a)\) |
Total energy for any Keplerian ellipse |
| Specific energy |
\(\varepsilon=v^2/2-GM/r\) |
Energy per unit mass |
| Escape speed |
\(v_{\mathrm{esc}}=\sqrt{2GM/r}\) |
Speed where \(E=0\) |
11. Assumptions and limits
- The model uses Newtonian gravity.
- The central body is treated as spherical or point-like.
- Atmospheric drag, thrust, and non-gravitational forces are ignored.
- The zero of gravitational potential energy is chosen at infinity.
- The formula \(E=-GMm/(2a)\) applies to ideal Keplerian bound orbits.
- Real spacecraft may need corrections for drag, thrust, oblateness, and perturbations from other bodies.
Key idea: orbital energy is negative for closed orbits because the object needs extra energy to escape to infinity.
